The probability interpretation of the wavefunction

The last chapter ended in an awkward place: de Broglie says particles have a wavelength, and the double slit says electrons really do interfere. So — what is waving?

In a water wave it is the height of the surface; in sound it is air pressure; in an electromagnetic wave it is the electric and magnetic fields. All of those can be measured directly. What about an electron wave? When Schrödinger wrote his equation in 1926 he had no answer either. He guessed that the wavefunction described the actual smeared-out distribution of the electron’s charge in space. That guess was disproved almost immediately.

The right answer came from Max Born, the same year, in a footnote.

The Born rule

Precisely: for a particle in the state $\Psi(x,t)$, one position measurement returns a result in $[a,b]$ with probability

$$P(a \le x \le b) = \int_a^b |\Psi(x,t)|^2 \,\dd x\tag{2.1.1}$$

Here $|\Psi|^2 = \Psi^*\Psi$, because $\Psi$ is generally complex. We call

$$\rho(x,t) \equiv |\Psi(x,t)|^2\tag{2.1.2}$$

the probability density. Watch its dimensions: in one dimension it is $1/\text{length}$, so $\rho$ itself is not a probability — $\rho\,\dd x$ is.

Why it has to be the modulus squared

Squaring the modulus of a complex number looks like a technicality. In fact it solves two problems at once.

Why |Ψ|², and not Ψ or Re Ψbasic~4 minexpand

A probability density has to satisfy two hard requirements:

1. Non-negativity: $\rho(x) \ge 0$. A probability cannot be negative.
2. Room for interference: when contributions from two paths are added, cancellation must be possible.

Could we use $\Psi$ directly? Solutions of the Schrödinger equation are generally complex, and $\Psi$ can be negative or even imaginary, so requirement 1 is dead on arrival.

What about $\operatorname{Re}\Psi$ or $|\Psi|$? Non-negativity can be forced by taking an absolute value, but that destroys requirement 2. The key point is that superposition happens at the level of $\Psi$, not at the level of $\rho$. In the double slit,

$$\Psi = \Psi_1 + \Psi_2\tag{2.1.5}$$

so that

$$\rho = |\Psi_1+\Psi_2|^2 = |\Psi_1|^2 + |\Psi_2|^2 + 2\operatorname{Re}(\Psi_1^*\Psi_2)\tag{2.1.6}$$

The last term is the interference term. In polar form $\Psi_j = |\Psi_j|\ee^{\ii\varphi_j}$ it equals

$$2|\Psi_1||\Psi_2|\cos(\varphi_2-\varphi_1)\tag{2.1.7}$$

and the phase difference decides whether it reinforces or cancels. If we added $|\Psi|$ directly, $\rho = (|\Psi_1|+|\Psi_2|)^2$ would always exceed the sum of the parts and fringes could never appear.

So: complex numbers supply the phase, the phase supplies interference, and the modulus squared supplies a non-negative probability. All three are needed. This is also why quantum mechanics cannot do without complex numbers — they are not a mathematical convenience but a physical necessity.

Have a look

The simulation below shows a particle in a one-dimensional infinite square well. The purple curve is $|\Psi|^2$, and the area under it over any interval is the probability of finding the particle there. The well itself is not solved until section 2.7; here only one thing matters: the blue curve ($\psi$) can be positive or negative, the purple one ($|\psi|^2$) never is.

◈The infinite square well

Drag the width L and the quantum number n and watch the energy, the waveform and the probability distribution move together. Units: ħ = m = 1.

Energy Eₙ (n=1)

4.935

relative to E₁(L=1)

1.00 ×

⟨x⟩ / L

0.500

Δx / L

0.180

Well width L1.00

Shrink the box and every level is pushed up together: E ∝ 1/L²

Quantum number n1

nodes = n − 1 = 0 (endpoints excluded)

State

Stationary ψₙSuperposition

Show ψShow |ψ|²

Turn the quantum number $n$ up to 3 and you will see $\psi$ cross zero twice inside the well. At those zeros the probability of finding the particle is exactly zero — even though there is substantial probability on both sides. A classical particle bouncing around a box could never manage to “never pass through some intermediate position”. That is already a strong quantum effect.

A concrete calculation

Worked example: probability over an intervalbasic~5 minexpand

Take a particle with wavefunction

$$\Psi(x) = \begin{cases} A\,x(L-x), & 0 \le x \le L\\ 0, & \text{otherwise}\end{cases}$$

Step 1: fix the normalisation constant $A$.

$$1 = \int_0^L |A|^2 x^2(L-x)^2\,\dd x = |A|^2\int_0^L (L^2x^2 - 2Lx^3 + x^4)\,\dd x\tag{2.1.8}$$

Term by term:

$$\int_0^L L^2x^2\dd x = \frac{L^5}{3},\quad \int_0^L 2Lx^3\dd x = \frac{L^5}{2},\quad \int_0^L x^4\dd x = \frac{L^5}{5}\tag{2.1.9}$$$$1 = |A|^2 L^5\left(\frac13-\frac12+\frac15\right) = |A|^2\frac{L^5}{30} \;\Longrightarrow\; A = \sqrt{\frac{30}{L^5}}\tag{2.1.10}$$

Step 2: find the probability in the left quarter.

$$P\left(0\le x\le \tfrac{L}{4}\right) = \frac{30}{L^5}\int_0^{L/4}x^2(L-x)^2\dd x\tag{2.1.11}$$$$= \frac{30}{L^5}\left[\frac{L^2x^3}{3}-\frac{Lx^4}{2}+\frac{x^5}{5}\right]_0^{L/4} = \frac{30}{L^5}\cdot L^5\left(\frac{1}{192}-\frac{1}{512}+\frac{1}{5120}\right)\tag{2.1.12}$$

Over the common denominator 15360: $\frac{80}{15360}-\frac{30}{15360}+\frac{3}{15360}=\frac{53}{15360}$, so

$$P = 30\cdot\frac{53}{15360} = \frac{53}{512}\approx 0.1035\tag{2.1.13}$$

About 10.4%. For comparison, a uniform distribution would give 25% over the same interval — this wavefunction pushes the particle towards the middle, because it vanishes at both ends.

∎Key formulas

• Born rule

  $$P(a\le x\le b)=\int_a^b|\Psi(x,t)|^2\,\dd x$$

  The only bridge between the wavefunction and experiment

• Probability density

  $$\rho(x,t)=|\Psi(x,t)|^2=\Psi^*\Psi\ \ge 0$$

  Dimensions of 1/length; ρ dx is the probability

• Interference term

  $$|\Psi_1+\Psi_2|^2=|\Psi_1|^2+|\Psi_2|^2+2|\Psi_1||\Psi_2|\cos(\varphi_2-\varphi_1)$$

  Superposition acts on Ψ; interference comes from the phase difference

?Self-check3 questions

1. 1.

   A wavefunction satisfies Ψ(x₀) = −0.4 (units suppressed). Which statement about the probability density near x₀ is correct?

   • The density is 0.16; the minus sign does not affect the probability
   • The density is −0.4, so there is negative probability there
   • The density is 0.4; you just take the absolute value
   • A negative wavefunction means the particle cannot be found there

   Submit
2. 2.

   Why must a quantum wavefunction be allowed to take complex values? (Select all that apply.)

   Select all that apply

   • A phase is needed to produce interference
   • Because probabilities have to be complex
   • A real non-negative function cannot satisfy non-negativity and allow interference at once
   • The Schrödinger equation contains i, so real solutions generally do not exist

   Submit
3. 3.

   For Ψ(x) = A·x(L−x) on 0 ≤ x ≤ L (zero elsewhere) with L = 1, find the probability of the particle lying in [0, 0.25] (three decimal places).

   3% relative tolerance

   Submit

What comes next

The condition $\int|\Psi|^2 = 1$ in the Born rule is not something imposed by hand — it has to hold at every instant, or particles would vanish into thin air. The next section proves that the Schrödinger equation guarantees exactly this, and picks up a quantity describing “where the probability is flowing” along the way.